If the temperature of a gas is increased by 1 K at constant pressure its volume increases by 0.0035 of the initial volume. The temperature of the gas is

Question

If the temperature of a gas is increased by 1 K at constant pressure its volume increases by 0.0035 of the initial volume. The temperature of the gas is (A) 100K (B) 150K (C) 300K (D) 285.7K

Tardigrade Admin · Accepted Answer

Let T be the initial temperature. So, T +1 be the final temperature. Let x = initial volume, then ( x +0.0035 x ) is the final volume. T 1 / T 2= V 1 / V 2 So, ( T / T +1)=( x / x +0.0035 x ) Or ( T +1/ T )=1.0035 Solving we get, T =285.7 K

Q. If the temperature of a gas is increased by 1 K at constant pressure its volume increases by 0.0035 of the initial volume. The temperature of the gas is

100K

150K

300K

285.7K

Let $T$ be the initial temperature. So, $T + 1$ be the final temperature.
Let $x =$ initial volume, then $(x + 0.0035 x)$ is the final volume.
$T_{1} / T_{2} = V_{1} / V_{2}$
So, $\frac{T}{T + 1} = \frac{x}{x + 0.0035 x}$
$O r \frac{T + 1}{T} = 1.0035$
Solving we get, $T = 285.7 K$

Q. If the temperature of a gas is increased by 1 K at constant pressure its volume increases by 0.0035 of the initial volume. The temperature of the gas is

100K

150K

300K

285.7K

Let T be the initial temperature. So, T+1 be the final temperature. Let x= initial volume, then (x+0.0035x) is the final volume. T1​/T2​=V1​/V2​ So, T+1T​=x+0.0035xx​ OrTT+1​=1.0035 Solving we get, T=285.7K

Let $T$ be the initial temperature. So, $T + 1$ be the final temperature.
Let $x =$ initial volume, then $(x + 0.0035 x)$ is the final volume.
$T_{1} / T_{2} = V_{1} / V_{2}$
So, $\frac{T}{T + 1} = \frac{x}{x + 0.0035 x}$
$O r \frac{T + 1}{T} = 1.0035$
Solving we get, $T = 285.7 K$